Answer:
[tex]\frac{d}{dx}\lbrack\frac{-5x^2+3x+2}{x}\rbrack=-\frac{5x^2+2}{x^2}[/tex]Step-by-step explanation:
To solve this, we'll use the quotient rule for derivatives. This rule tells us that:
[tex]\frac{d}{dx}\lbrack\frac{f(x)}{g(x)}\rbrack=\frac{g(x)f^{\prime}(x)-f(x)g^{\prime}(x)}{\lbrack g(x)\rbrack^2}[/tex]For the given expression, we'll have that:
[tex]\begin{gathered} f(x)=-5x^2+3x+2 \\ f^{\prime}(x)=-10x+3 \\ \\ g(x)=x \\ g^{\prime}(x)=1 \\ \lbrack g(x)\rbrack^2=x^2 \end{gathered}[/tex]This way, we'll have that:
[tex]\frac{d}{dx}\lbrack\frac{-5x^2+3x+2}{x}\rbrack=\frac{(x)(-10x+3)-(-5x^2+3x+2)(1)}{x^2}[/tex]Simplifying this expression,
[tex]\begin{gathered} \frac{(x)(-10x+3)-(-5x^{2}+3x+2)(1)}{x^{2}} \\ \\ \rightarrow\frac{-10x^2+3x-+5x^2-3x-2}{x^2} \\ \\ \rightarrow\frac{-5x^2-2}{x^2} \\ \\ \Rightarrow-\frac{5x^2+2}{x^2} \end{gathered}[/tex]Thus, we can conclude that:
[tex]\frac{d}{dx}\lbrack\frac{-5x^2+3x+2}{x}\rbrack=-\frac{5x^2+2}{x^2}[/tex]
